DOUBLING TOLERANCES AND COALITION LATTICES

GABOR´ CZEDLI´

Dedicated to the memory of Ivo G. Rosenberg

Abstract. If every block of a (compatible) tolerance (relation) T on a mod- ular L of finite length consists of at most two elements, then we call T a doubling tolerance on L. We prove that, in this case, L and T determine a modular lattice of size 2|L|. This construction preserves distributivity and modularity. In order to give an application of the new construct, let P be a (poset). Following a 1995 paper by G. Poll´akand the present author, the of P are called the coalitions of P . For coalitions X and Y of P , let X ≤ Y mean that there exists an injective map f from X to Y such that x ≤ f(x) for every x ∈ X. If P is a finite chain, then its coalitions form a by the 1995 paper; we give a new proof of the distributivity of this lattice by means of doubling tolerances.

1. Introduction There are two words in the title that are in connection with Ivo G. Rosenberg; these words are “tolerance” and “lattice”, both occurring also in the title of our joint lattice theoretical paper [6] (coauthored also by I. Chajda). This fact encouraged me to submit the present paper to a special volume dedicated to Ivo’s memory even if this volume does not focus on lattice theory. The paper is structured as follows. In Section 2, after few historical comments on lattice tolerances, we introduce the concept of doubling tolerances (on lattices) as those tolerances whose blocks are at most two-element. We prove that finite modular lattices can be “doubled” with the help of this tolerances; see Theorem 2.1. Furthermore, this doubling construction preserves modularity and distributivity. In Section 3, after recalling the concept of coalition lattices of certain finite posets (partially ordered sets) and some related results and after presenting some new observations, we use our doubling construction to give a new proof of the fact that coalition lattices of finite chains are distributive; see Lemma 3.4. Also, this new

arXiv:1909.09539v2 [math.RA] 10 Dec 2019 proof provides a natural example of doubling tolerances and our construct. Note in advance that every structure in this paper is assumed to be of finite length even if this is not always emphasized.

2. Doubling tolerances Tolerances (that is, compatible tolerance relations) of lattices were first inves- tigated by Chajda and Zelinka [7]. They are reflexive and symmetric relations

Date: Always check the author’s homepage for updates! December 10, 2019. 1991 Mathematics Subject Classification. 006B99, 06C99, 06D99. Key words and phrases. Lattice tolerance, modular lattice, coalition lattice. This research was supported by NFSR of Hungary (OTKA), grant number K 115518. 1 2 G. CZEDLI´ preserved by both lattice operations. Tolerances on lattices have been studied for long; see, for example Bandelt [2], Chajda [4], Cz´edli[8], Cz´edliand Gr¨atzer[11], Grygiel and Radeleczki [16], and Kindermann [19]. Apparently, apart from some artificial constructs like those in Chajda, Cz´edli,and Halaˇs[5], tolerances seem to be interesting only in lattices and lattice-like structures. A lattice L is of finite length if there is a natural number n such that no chain in L has more than n + 1 elements; if so then the least such n is the length of L. For a tolerance T on a lattice L of finite length, maximal subsets X of L such that X × X ⊆ T are called the blocks of T ; they are known to be intervals; see, for example, Cz´edli [8]. By [8, Proposition 1] and its dual, for arbitrary blocks [a, b] and [c, d] of T , a = c ⇐⇒ b = d. (2.1) If L (of finite length as always) is modular and no block of T has more than two elements or, equivalently, if the “covering or equal” relation a  b holds for every block [a, b] of T , then T will be called a doubling tolerance on L. Let us emphasize that we define doubling tolerances only on modular lattices of finite lengtht. If T is a doubling tolerance on L, then its intersection with the covering relation ≺ (or, equivalently, with the strict lattice ordering <) will be denoted by ≺T . That is, a ≺T b ⇐⇒ [a, b] is a two-element block of T . Note that ≺T determines T . We will also need the negated relation: a 6≺T b will mean that a ≺T b fails. The two-element chain will be denoted by C2 := {0, 1}; with 0 ≺ 1, of course. The direct product order on L × C2 will be denoted by ≤ or π. That is, each of (a, i) ≤ (b, j), (a, i) π (b, j), and ((a, i), (b, j)) ∈ π means that a ≤ b in L and i ≤ j in C2. On L × C2, we define a relation τ as follows: def (a, i) τ (b, j) ⇐⇒ i = 1, j = 0 and a ≺T b. We will also need the negated relation:

(a, i) τ (b, j) ⇐⇒def (a, i) τ (b, j) does not hold.

Let ρ the transitive closure of π ∪ τ; it is a relation on L × C2. Finally, we denote the structure (L × C2; ρ) by [L ∗2 T ]. The subscript of ∗ reminds us that |L × C2| = 2 · |L|, so what we do is doubling in some sense. With the concepts and notations introduced above, we are now in the position to formulate our main result.

Figure 1. Two examples; the T -blocks are the dotted ovals DOUBLING TOLERANCES AND COALITION LATTICES 3

Theorem 2.1. If T is a doubling tolerance on a modular lattice L of finite length, then [L ∗2 T ] = (L × C2; ρ) is also a modular lattice of finite length. If, in addition, L is distributive, then so is [L ∗2 T ]. Remark 2.2. If T is the equality relation, all of whose blocks are singletons, then [L ∗2 T ] is simply the direct product of L and C2. Some examples where T is not the equality relation are given in Figures 1 and Figure 2.

Figure 2. Two further examples; the lower elements are black 0 ∼ filled and [CoalL(P ) ∗2 T ] = CoalL(P)

Proof of Theorem 2.1. Let T be a doubling tolerance on a modular lattice L of finite length. The elements of L × {0} will be called lower elements while those of L × {1} are the upper elements. Observe that, for any (a, i), (b, j) ∈ L × C2, if (a, i) ρ (b, j), then a ≤ b in L; (2.2) simply because both π and τ have the same property. By a (π ∪ τ)-sequence from (a, i) ∈ L × C2 to (b, j) ∈ L × C2 we mean a finite sequence

(a, i) = (x0, k0) π∪τ (x1, k1) π∪τ . . . π∪τ (xn, kn) = (b, j) (2.3) of elements of L × C2. By the definition of ρ, we have that (a, i) ρ (b, j) if and only if there is a sequence described in (2.3). In order to show that ρ is antisymmetric, assume that (a, i) ρ (b, j) and (b, j) ρ (a, i). By (2.2), a = b, whereby the only way of violating antisymmetry is that (a, 1) ρ (a, 0). Take a (π ∪ τ)-sequence from (a, 1) to (a, 0); see (2.3). Since only a “τ-step” in this sequence can change an upper element to a lower element, at least one τ-step occurs in this sequence. But this step strictly increases the first component, whereby it follows from (2.2) that the first component cannot remain a at the end of the sequence. Thus, (a, 1) ρ (a, 0) is impossible and ρ is antisymmetric. The reflexivity of ρ follows from ρ ⊆ π while ρ is transitive by its definition. Hence, [L ∗2 T ] is a poset; note that the modularity of L has not yet been used. Now that we know that ρ is a partial ordering, we can speak about the corre- sponding covering relation, which will be denoted by ≺ρ. Self-explanatory analo- gous notations, like <ρ, ≤ρ, or ≺π will also be used. If the superscript is dropped, then the meaning should be clear for the context since ≤, ≺, < refer to the original ordering of L when they are applied for the elements of L, and they refer to ρ between elements of [L ∗2 T ]. We claim that, for arbitrary (a, i), (b, j) ∈ L × C2, we have that (a, i) ≺ρ (b, j) if and only if one of the following three possibilities holds. (cov-1) a = b, i = 0, and j = 1. (cov-2) i = j, a ≺ b in L, and a 6≺T b ( that is, [a, b] is not a T -block). (cov-3) i = 1, j = 0, and a ≺T b. 4 G. CZEDLI´

In order to verify that the disjunction of (cov-1), (cov-2), and (cov-3) describes ≺ρ correctly, assume that (a, i) ≺ρ (b, j). Take a repetition-free (π ∪ τ)-sequence from (a, i) ∈ L × C2 to (b, j); see (2.3). We can assume that every π-step in this sequence is a ≺π-step. Clearly, the sequence consists of a single step, which is either a τ-step, corresponding to (cov-3), or it is a ≺π-step (a, 0) ≺π (a, 1) or (a, i) ≺π (b, i) with a ≺ b in L; furthermore, if (a, i) ≺π (b, i) with a ≺ b in L, then a 6≺T b since otherwise (a, 1) τ (b, 0) and (a, 0) <ρ (a, 1) <ρ (b, 0) (for i = j = 0) or (a, 1) <ρ (b, 0) <ρ (b, 1) (for i = j = 1) would contradict (a, i) ≺ρ (b, j). So whenever (a, i) ≺ρ (b, j), then at least one of (and exactly one of) (cov-1), (cov-2), and (cov-3) holds. Since a τ-step of a (π∪τ)-sequence increases the first component, it follows that (cov-1), which a special sort of a π-covering, is indeed a ρ-covering. Since a 6≺T b in (cov-2), it follows from (2.2) that no τ-covering step can interfere and the π-covering described in (cov-2) is a ρ-covering, as required. Finally, since ρ is antisymmetric, we have that for every x ∈ L,

(x, 1) 6≤ρ (x, 0). (2.4)

Now if (cov-3) holds for (a, i) and (b, j), then a ≺T b yields that a ≺ b while the definition of τ and ρ implies that (a, i) = (a, 1) <ρ (b, 0) = (b, j). Take an arbitrary (c, k) ∈ L × C2 such that (a, 1) ≤ρ (c, k) ≤ρ (b, 0). Since (a, 1) ≤π (c, k) ≤π (b, 0) by (2.2) and a ≺ b, we have that c ∈ {a, b}. If c = a, then we can apply (2.4) with x := a to exclude that k = 0, whence (c, k) = (a, 1). Similarly, if c = b, then (2.4) with x := b excludes that k = 1, whereby (c, k) = (b, 0). Thus (c, k) is necessarily (a, 1) or (b, 0), and it follows that (a, i) = (a, 1) ≺ρ (b, 0) = (b, j). Hence, (cov-3) describes a ρ-covering, as required. We have seen that the disjunction of (cov-1), (cov-2), and (cov-3) describes ≺ρ correctly. Next, for later use, we observe the following. For covering pairs e1 ≺ f1 and e2 ≺ f2 in L, we say that they are transposed if e2 ∨ f1 = f2 and e2 ∧ f1 = e1, or e1 ∨ f2 = f1 and e1 ∧ f2 = e2. Since T is compatible, if e ≺ f and e ≺ f are transposed edges  1 1 2 2 (2.5) in L, then e1 ≺T f1 ⇐⇒ e2 ≺T ef ; referencing this property is how we can exploit the compatibility of T . Next, (2.6) formulates the BEZ Lemma, named after the initials of its inventors, Bj¨orner,Edelman, and Ziegler [3, Lemma 2.1]. As a preparation to it, recall that a poset is bounded if it has a (necessarily unique) least element 0 and a (necessarily unique) greatest element 1, and a poset is of finite length is there is a finite upper bound on the lengths of its chains. If P is a bounded poset of finite length such that for ) any x and y in P with a common lower cover the join (2.6) x ∨ y exists, then P is a lattice.

Since π ⊆ ρ, it is clear that [L ∗2 T ] is a bounded poset with bottom element (0, 0) and top element (1, 1). Let

... (a−1, k−1) <ρ (a0, k0) <ρ (a1, k1) <ρ ... (2.7) be an arbitrary chain in [L ∗2 T ]. Since L is of finite length, say, of length n, it follows from (2.2) that the set {. . . , a−1, a0, a1,... } consists of at most n + 1 elements. The set {. . . , k−1, k0, k1,... } has at most two elements since it is a of C2. Hence, the chain in (2.7) consists of at most 2(n + 1) elements, and we obtain that [L ∗2 T ] is of finite length. Hence, the BEZ Lemma is applicable. In DOUBLING TOLERANCES AND COALITION LATTICES 5 fact, for later use, we are going to prove a bit more than required by (2.6). By a covering square in a poset we mean a quadruple (o, a, b, i) of four distinct elements such that o ≺ a, o ≺ b, a ≺ i, and b ≺ i. We claim that  If x, y, z ∈ L × C2 such that x 6= y, z ≺ρ x, and  z ≺ρ y, then the join x ∨ y exists in [L ∗2 T ] and (2.8) (z, x, y, x ∨ y) is a covering square in [L ∗2 T ].  In order to prove (2.8), we have to deal with several cases depending on the position of z and the covering types (cov-1),. . . , (cov-3) that occur.

Case 1. We assume that z is of the form (c, 0) and both of z ≺ρ x and z ≺ρ y are (cov-2)-coverings. Then x = (a, 0), y = (b, 0), c ≺ a, c ≺ b, c 6≺T a, and c 6≺T b. Letting u := (a ∨ b, 0), we claim that u = x ∨ y and (z, x, y, u) is a covering square in [L ∗2 T ]. By the modularity of L, a ≺ a ∨ b and b ≺ a ∨ b in L. So, (c, a, b, a∨b) is a covering square in L. By (2.5), a 6≺T a∨b and b 6≺T a∨b, whereby x = (a, 0) ≺ (a ∨ b, 0) and y = (b, 0) ≺ (a ∨ b, 0) are (cov-2)-coverings, so we have a covering square in [L ∗2 T ]. Clearly, (a ∨ b, 0) is an upper bound of x = (a, 0) and y = (b, 0), and it follows easily from (2.2) that is the least upper bound. Hence, x ∨ y = (a ∨ b, 0), so the required join exists. This completes Case 1. Before the next case, we prove the following auxiliary statement. If u ≺ v, u 6≺ v, and v ≤ w in L, then T (2.9) (u, 1) ρ (w, 0) implies that (v, 1) ρ (w, 0). In order to show this, take a shortest repetition-free (π ∪ τ)-sequence from (u, 1) to (w, 0). As in (2.3), let (xi, ki), i = 0, 1, . . . , n, be the members of this sequence. Since (x0, k0) = (u, 1) is an upper element but (xn, kn) = (w, 0) is not, there is a unique integer t ∈ {0, 1, . . . , n − 1} such that the (xi, ki) are upper elements for i = 0, 1, . . . , t but (xt+1, kt+1) is a lower element. That is, k0 = k1 = . . . kt = 1 but kt+1 = 0. If there is an i ∈ {0, . . . , t} such that v ≤ xi, then (v, 1) ≤ρ (xi, 1) by π ⊆ ρ,(xi, 1) ≤ρ (w, 0) by the second half of the sequence, and so the transitivity of ρ yields that (v, 1) ρ (w, 0), as required. Hence, we can assume that v 6≤ xi for i = 0, 1, . . . , t. The transition from (xt, kt) = (xt, 1) to (xt+1, kt+1) = (xt+1, 0) in the sequence is not a π-step, whence it is a τ-step. This implies that xt ≺T xt+1 and, in particular, xt ≺ xt+1. Now there are two cases depending on whether v ≤ xt+1 or not. First, let v ≤ xt+1. Since v 6≤ xt and xt ≺ xt+1, it follows from xt < xt ∨v ≤ xt+1 that xt ∨ v = xt+1. Similarly, u ≤ xt by (2.2), and so u ≤ xt ∧ v < v and u ≺ v yield that xt ∧ v = u. That is, xt ≺ xt+1 and u ≺ v are transposed edges. But this contradicts (2.5) since xt ≺T xt+1 but u 6≺T v. Second, assume that v 6≤ xt+1. Since u ≤ xt+1 by (2.2), u ≤ xt+1 ∧ v < v and u ≺ v give that xt+1 ∧ v = u. By the Isomorphism Theorem for Modular Lattices, see, for example, Gr¨atzer[14,Theorem 348], the maps ) ϕ:[u = xt+1 ∧ v, xt+1] → [v, xt+1 ∨ v], defined by s 7→ s ∨ v, and (2.10) ψ :[v, xt+1 ∨ v] → [u, xt+1], defined by s 7→ s ∧ xt+1 are reciprocal lattice isomorphisms. Hence, using that xt ≺ xt+1, we have that xt ∨ v = ϕ(xt) ≺ ϕ(xt+1) = xt+1 ∨ v. Furthermore, xt+1 ∨ xt ∨ v = xt+1 ∨ v and xt = ψ(ϕ(xt)) = (xt ∨ v) ∧ xt+1, showing that xt ≺ xt+1 and xt ∨ v ≺ xt+1 ∨ v are transposed edges. This fact, xt ≺T xt+1, and (2.5) give that xt ∨ v ≺T xt+1 ∨ v. 6 G. CZEDLI´

Hence, (xt ∨ v, 1) τ (xt+1 ∨ v, 0), which gives that (xt ∨ v, 1) ≤ρ (xt+1 ∨ v, 0). Since π ⊆ ρ, we have that (v, 1) ≤ρ (xt ∨ v, 1). Since (xt+1, 0) occurs in our sequence, xt+1 ≤ w by (2.2). Hence, using the assumption v ≤ w, we have that xt+1 ∨ v ≤ w and we conclude from π ⊆ ρ that (xt+1 ∨ v, 0) ≤ρ (w, 0). The last three inequalities with ≤ρ yield that (v, 1) ≤ρ (w, 0), in other words, (v, 1) ρ (w, 0), proving (2.9). Note that although the use of modularity could have been avoided at several places in our considerations, it seems to be important at (2.10).

Case 2. We assume that z = (c, 0) is a lower element and at least one of z ≺ρ x and z ≺ρ y is a (cov-1)-covering. The other covering is necessarily a (cov-2)-covering, so we can assume that x = (c, 1) and y = (b, 0) with c ≺ b and c 6≺T b. We claim that (b, 1) is the join of x and y; it is clearly an upper bound. Let (d, k) be an arbitrary upper bound of x = (c, 1) and y = (b, 0). By (2.2), b ≤ d. If k = 1, then (b, 1) ≤ρ (d, 1) = (d, k) by π ⊆ ρ, as required. Thus, we assume that k = 0. Letting (c, b, d) play the role of (u, v, w), we obtain from (2.9) that (b, 1) ≤ρ (d, 0) = (d, k), showing that (b, 1) is the join of x = (c, 1) and y = (b, 0) in [L ∗2 T ]. Since c 6≺T b and c ≺ b, we have that x = (c, 1) ≺ρ (b, 1) is a (cov-2)- covering. Since y = (b, 0) ≺ρ (b, 1) is a (cov-1)-covering, (z, x, y, x ∨ y) is a covering square in [L ∗2 T ], completing Case 2.

Case 3. We assume that z = (c, 1) is an upper element and at least one of the coverings z ≺ρ x and z ≺ρ y is a (cov-3)-covering. Let z ≺ρ x such a covering, that is, x = (a, 0) such that c ≺T a (and so c ≺ a). We obtain from (2.1) and y 6= x that z ≺ρ y cannot be a (cov-3)-covering, whereby it is a (cov-2) covering, that is, y = (b, 1) with c ≺ b and c 6≺T b. By the (upper semi-)modularity of L,(c, a, b, a∨b) is a covering square in L. This fact and (2.5) give that b ≺T a ∨ b but a 6≺T a ∨ b. Thus, y = (b, 1) ≺ρ (a ∨ b, 0) is a (cov-3)-covering and x = (a, 0) ≺ρ (a ∨ b, 0) is a (cov-2)-covering. We have seen that (z, x, y, (a ∨ b, 0)) is a covering square in [L ∗2 T ]. In particular, (a ∨ b, 0) is an upper bound of x = (a, 0) and y = (b, 1). Let (d, k) be another upper bound. Since a ∨ b ≤ d by (2.2), we obtain from π ⊆ ρ that (a ∨ b, 0) ≤ (d, k), as required. This completes Case 3.

Case 4. We assume that z = (c, 1) is an upper element and none of the coverings z ≺ρ x and z ≺ρ y is a (cov-3)-covering. Then both are (cov-2)-coverings, so x = (a, 1), y = (b, 1), c ≺ a, c ≺ b, c 6≺T a, and c 6≺T b. By the modularity of L, (c, a, b, a ∨ b) is a covering square, and it follows from (2.5) that a 6≺T a ∨ b and b 6≺T a ∨ b. Hence, (z, x, y, (a ∨ b, 1)) is a covering square in [L ∗2 T ] with all of its edges being (cov-2)-coverings. This square shows that (a ∨ b, 1) is an upper bound of x and y. Let (d, k) be an arbitrary upper bound of x = (a, 1) and y = (b, 1) in [L ∗2 T ]. From (2.2), it follows that a ∨ b ≤ d. Observe that π ⊆ ρ gives that (a ∨ b, 1) ≤ρ (d, 1). So we can assume that k = 0 since otherwise (a ∨ b, 1) ≤ρ (d, k) has already been shown. Since (a, 1) ≤ρ (d, 0), a ≺ a ∨ b, a 6≺T a ∨ b, and a ∨ b ≤ d, we can apply (2.9) with (a, a ∨ b, d) playing the role of (u, v, w) to conclude that (a ∨ b, 1) ≤ρ (d, 0) = (d, k). This completes Case 4.

Cases 1–4 prove the validity of (2.8), and so [L ∗2 T ] is a lattice by (2.6). The following statement will be used to prove that [L ∗2 T ] is modular. We claim that

If x, y, z ∈ L × C such that x 6= y, x ≺ z, and y ≺ z,  2 ρ ρ (2.11) then (x ∧ y, x, y, z) is a covering square in [L ∗2 T ]. DOUBLING TOLERANCES AND COALITION LATTICES 7

Now that we already know that [L ∗2 T ] is a lattice, the proof of (2.11) is easier than that of (2.8). Indeed, it suffices to show that x and y from (2.11) have a common lower cover. However, we have to deal with several cases again. First, assume that at least one x and y, let it be x, is (cov-3)-covered by z. Hence z = (c, 0), x = (a, 1), and a ≺T c (and so a ≺ c). By (2.1), x is the only lower (cov-3)-cover of x, whereby y ≺ρ z is a (cov-2)-covering, y is of the form y = (b, 0) with b ≺ c and b 6≺T c. By the modularity of L,(a ∧ b, a, b, c) is a covering square in L. Hence, using (2.5), we obtain the a ∧ b ≺T b and a ∧ b 6≺T a. Therefore, (a ∧ b, 1) ≺ρ (b, 0) = y is a (cov-3)-covering while (a ∧ b, 1) ≺ρ (a, 1) = x is a (cov-2)-covering, showing that x and y have a common lower cover, as required. Thus, in the rest of the cases, we can disregard the situation when x ≺ρ z or y ≺ρ z is a (cov-3)-covering. Second, assume that both x ≺ z and y ≺ z are (cov-2)-coverings. Then z = (c, i), x = (a, i), y = (b, i), a ≺ c, b ≺ c, a 6≺T c, and b 6≺T c. The modularity of L yields that (a ∧ b, a, b, c) is a covering square in L. Hence, a ∧ b 6≺T a and a ∧ b 6≺T b by (2.5), and it follows that (a∧b, i) is (cov-2)-covered both by x = (a, i) and y = (b, i). So x and y has a common lower cover in this case. Third, since no element can have two distinct lower (cov-1)-covers, there remains only one case: one of x ≺ρ z and y ≺ρ z is a (cov-1)-covering and the other one is a (cov-2)-covering. Hence, we can assume that z = (c, 1), x = (a, 1) with a ≺ c but a 6≺T c, and y = (c, 0). Clearly, (a, 0) ≺ρ (a, 1) = x is a (cov-1)-covering while (a, 0) ≺ρ (c, 0) = y is a (cov-2)-covering, showing that x and y have a common lower cover again. This proves (2.11).

Figure 3. The lattices in Jakub´ık’stheorem; see (2.12)

If L1 is a sublattice of another lattice L2 such that every covering pair a ≺L1 b is also a covering pair in L2, then L1 is a cover-preserving sublattice of L2. However, instead of saying that M3 is cover-preserving sublattice of L, we usually say shortly 0 that L has a covering M3. The lattices B, B , M3, and L(m, n) (for m ≥ 3 and n ≥ 4) are given in Figure 3. It is proved in Jakub´ık[17, Theorems 1 and 2] that for a lattice L of finite length (in particular, for a finite lattice L), L is modular if and only if none of the lattices B,  B0, and L(m, n)(m ≥ 3, n ≥ 4) is a cover-preserving  (2.12) sublattice of L. Furthermore, L is distributive if and  only if it is modular and it has no covering M3. Observe that each of the L(m, n)(m ≥ 3, n ≥ 4) and B has three elements, x, y, z, such that x ≺ z, y ≺ z, x 6= y, but (x ∧ y, x, y, z) is not a covering square. It follows from (2.11) that these lattices cannot be cover-preserving sublattices of [L ∗2 T ]. Similarly, B0, the dual of B, has elements x, y, z such that z ≺ x, z ≺ y, x 6= y, but (z, x, y, x ∨ y) is not a covering square in B0. Hence, B0 is not a cover-preserving sublattice of [L ∗2 T ] by (2.8). Therefore, (2.12) yields that [L ∗2 T ] is modular. 8 G. CZEDLI´

Before dealing with the distributive case, the following auxiliary statement is worth proving.  If (d1, i) ≺ρ (d2, i) and (d1, i) ≺ρ (d3, i)  are (cov-2)-coverings, then (d2, i) ∨ (d3, i) = (2.13) (d2 ∨ d3, i) holds in [L ∗2 T ]. 

Indeed, if the premise of (2.13) holds, then d1 ≺ d2 and d1 ≺ d3, so modularity yields that (d1, d2, d3, d2 ∨ d3) is a covering square in L. Since d1 6≺T d2 and d1 6≺T d3, we obtain from (2.5) that d2 6≺T d2 ∨ d3 and d3 6≺T d2 ∨ d3. Hence, (d2, i) ≺ρ (d2 ∨ d3, i) and (d3, i) ≺ρ (d2 ∨ d3, i), since they are (cov-2)-coverings. We are in a lattice, so these two coverings yield the validity of (2.13). Next, assume that L is distributive; we already know that [L ∗2 T ] is modular. For the sake of contradiction, suppose that [L ∗2 T ] is not distributive. By (2.12), [L ∗2 T ] has a covering M3 = {u, x, y, z, v}, where u ≺ρ x ≺ρ v, u ≺ρ y ≺ρ v, and u ≺ρ z ≺ρ v. There are three cases to consider. First, assume that each of u ≺ρ x, u ≺ρ y, and u ≺ρ z is a (cov-2)-covering. Then we can write that u = (e, i), x = (a, i), y = (b, i), and z = (c, i), and it follows from (2.13) that each of (a∨b, i), (a∨c, i) and (b∨c, i) equals v. Hence, a∨b = a∨c = b∨c, yielding that (e, a, b, c, a∨b) is a (covering) M3 in L, contradicting the distributivity of L. Second, assume that least one of u ≺ρ x, u ≺ρ y, and u ≺ρ z is a (cov-1)- covering. Then u is a lower element of the form u = (e, 0), whence none of u ≺ρ x, u ≺ρ y, and u ≺ρ z is a (cov-3)-covering. Since at most one of these three coverings can be a (cov-1)-covering, we can assume that u = (e, 0) ≺ρ (e, 1) = x is a (cov-1)- covering while u ≺ρ (b, 0) = y and u ≺ρ (c, 0) = z are (cov-2)-coverings. By (2.13), v = y ∨ z = (b ∨ c, 0). Hence, (e, 1) = x ≺ρ v = (b ∨ c, 0) is a covering, because we are in a covering M3. The only way that a lower element can cover an upper one is a (cov-3)-covering. Hence, (e, 1) ≺ρ (b ∨ c, 0) is a (cov-3)-covering, e ≺T b ∨ c and, in particular, e ≺ b ∨ c. But this is a contradiction, because the (cov-2)-coverings (e, 0) = u ≺ρ y = (b, 0) and (b, 0) = y ≺ρ v = (b ∨ c, 0) give that e < b < b ∨ c. Third, assume that least one of u ≺ρ x, u ≺ρ y, and u ≺ρ z is a (cov-3)-covering. Then u is an upper element of the form u = (e, 1); let u ≺ρ x a (cov-3)-covering. Then x is of the form x = (a, 0) with e ≺T a and, in particular, e ≺ a. It follows from (2.1) that none of u ≺ρ y, and u ≺ρ z is a (cov-3)-covering, and they are not (cov-1)-coverings because u is an upper element. So u ≺ρ y and u ≺ρ z are (cov-2)- coverings and we can write that y = (b, 1) and z = (c, 1) with b 6= c. From (2.13), we obtain that v = y ∨ z = (b ∨ c, 1). Hence, (a, 0) = x ≺ρ v = (b ∨ c, 1). This is neither a (cov-2)-covering, nor a (cov-3)-covering, because x is a lower element and v is an upper one. Thus, (a, 0) ≺ρ (b ∨ c, 1) is a (cov-1)-covering and so a = b ∨ c. Hence, e < b < b ∨ c = a in L, contradicting e ≺ a. Now that all the three cases have led to contradiction, we have shown that the modular lattice [L ∗2 T ] has no covering M3, and it follows from (2.12) that this lattice is distributive. The proof of Theorem 2.1 is complete. 

3. An application of doubling tolerances to coalition lattices For a finite poset P = (P ; <), the set of all subsets of P will be denoted by Pow(P ) or Pow(P). Note at this point that the relations < and ≤ in a poset P DOUBLING TOLERANCES AND COALITION LATTICES 9 mutually determine each other; this allows us to use both of them even if only one is given originally. Definition 3.1 (Cz´edliand Poll´ak[12]). Let P = (P ; <) be a finite poset. For X,Y ∈ Pow(P), a map (function) ϕ: X → Y is extensive if ϕ is injective and x ≤ ϕ(x) holds for every x ∈ X. Let X ≤ Y mean that there exists an extensive map X → Y . With this meaning of “≤”, the poset CoalL(P) = (Pow(P); ≤) is the coalition poset of P, and its elements are called the coalitions of P. When we consider a subset X of P as a member of CoalL(P), then we call it a coalition of P rather than a subset. Note that since 1995, when [12] was published, the terms “coalition” and “coalition lattice” have also been used with different meanings in mathematics and informatics; see, e.g., [1], [18], and [20]. The Hasse diagram of our finite poset P = (P ; <) is also a graph; the (connec- tivity) components of this graph are the components of P. These components are also posets with the orderings restricted from P to them. We say that P is upper bound free if no two incomparable elements has an upper bound in P. Lower bound free posets are defined dually. Note that P is both upper bound free and lower bound free if and only if all of its components are chains. In [12], we proved that Proposition 3.2 (Cz´edliand Poll´ak[12]). Let P = (P ; <) be a finite poset. (i) CoalL(P) is a lattice if and only if P is upper bound free. (ii) If P1,..., Pn is the list of the components of P, then the lattice CoalL(P) Qk is (isomorphic to) the direct product i=1 CoalL(Pi) (iii) If CoalL(P) is a lattice and P is lower bound free, then CoalL(P) is dis- tributive. (iv) If CoalL(P) is a distributive lattice, then P is lower bound free. Next, we formulate a particular case of 3.2(iii), which we are going to prove here with the help of doubling tolerances; note that the conjunction of this particular case with (the more or less trivial) 3.2(ii) implies 3.2(iii). Corollary 3.3. If P = (P ; <) is a finite chain, then CoalL(P) is a distributive lattice. We call the statement above a corollary because there will be no separate proof of it; this corollary will prompt follow from the following lemma, the main achievement of (the current) Section 3. The primary purpose of this lemma is to present an example and an application of our doubling construction. Lemma 3.4. Let P = (P ; <) be a finite non-singleton chain with smallest element 0, and let w denote its unique . Also, let P0 be its principal filter ↑w; that is, P 0 = P \{0}. On the lattice CoalL(P0), we define a relation T as follows: T := {(A, B) ∈ CoalL(P0)2 : A ∪ {w} = B ∪ {w}}. (3.1) 0 0 Then T is a doubling tolerance on CoalL(P ) and [CoalL(P ) ∗2 T ] is isomorphic to CoalL(P). Also, T is a congruence and both CoalL(P0) and CoalL(P) are dis- tributive lattices. Next, we state and prove some lemmas that will be needed in the proof of Lemma 3.4; these lemmas can be of separate interest.

Lemma 3.5. If P is a finite chain and A1,A2 ∈ CoalL(P), then

A1 ∩ A2 ⊆ A1 ∧ A2 ⊆ A1 ∪ A2 and A1 ∩ A2 ⊆ A1 ∨ A2 ⊆ A1 ∪ A2. 10 G. CZEDLI´

Proof. First, with the notation given in Lemma 3.5, let A1,A2 ∈ CoalL(P) be nonempty coalitions. Let c be the largest element of the nonempty set

H := {x1 ∧ x2 : x1 ∈ A1 and x2 ∈ A2}.

Since P is a chain, c exists, H ⊆ A1 ∪ A2, and, in particular, c ∈ A1 ∪ A2. Pick b1 ∈ A1 and b2 ∈ A2 such that c = b1 ∧b2. For i ∈ {1, 2}, let ai = c if c ∈ Ai, and let 0 ai = bi otherwise. Then a1 ∈ A1, a2 ∈ A2, and c = a1 ∧ a2. Let Ai := Ai \{ai} for = 1, 2, and let Pb := P \{c}. Then (Pb; <) =: Pb is a subchain of P. For i ∈ {1, 2}, 0 0 we have that c∈ / Ai, since otherwise c ∈ Ai ⊆ Ai would give that ai = c and so 0 0 0 ai = c ∈ Ai = Ai \{ai} would be a contradiction. Hence, and A1,A2 ∈ CoalL(Pb). 0 0 0 Let C be the meet of A1 and A2 in CoalL(Pb). As a particular case of Cz´edliand Poll´ak[12, Proposition 1], A ∧ A = C0 ∪ {c}, that is, A ∧ A = (A0 ∧ A0 ) ∪ {c}. (3.2) 1 2 1 CoalL(P) 2 1 CoalL(Pb) 2 Next, we claim that

for any A1,A2 ∈ CoalL(P), we have that A1 ∧ A2 ⊆ A1 ∪ A2. (3.3)

We prove this by induction on |P |. If P is a singleton, then (3.3) is clear. If A1 or A2 is the empty coalition, then so is A1 ∧ A2 and (3.3) is clear again. So, for the induction step, we can assume that |P | > 1, (3.3) holds for smaller chains, and none of A1 and A2 is empty. With the notation used in (3.2), c ∈ A1 ∪ A2. By the induction hypothesis, C0 = A0 ∧ A0 ⊆ A0 ∪ A0 ⊆ A ∪ A . Hence, (3.2) 1 CoalL(Pb) 2 1 2 1 2 implies that A1 ∧ A2 ⊆ A1 ∪ A2, proving (3.3). Since the map δ : CoalL(P) → CoalL(P), defined by A 7→ P \ A, is a dual lattice automorphism by Cz´edliand Poll´ak[12, Proposition 2] and δ is an involution, we obtain that A1 ∨A2 = δ(δ(A1)∧δ(A2)). Note that δ is also a dual automorphism of the powerset lattice (Pow(P ); ⊆) by the de Morgan laws. Hence, letting Xi = δ(Ai), applying (3.3) for X1 and X2, and using that δ is an involution, we obtain that for any A1,A2 ⊆ CoalL(P), (3.3) A1 ∨ A2 = δ(X1 ∧ X2) ⊇ δ(X1 ∪ X2) = δ(X1) ∩ δ(X2) = A1 ∩ A2. (3.4)

It has been proved in Cz´edli[9, displays in page 102] that A1 ∩ A2 ⊆ A1 ∧ A2 and A2 ∨ A2 ⊆ A1 ∪ A2, even without assuming that P is a chain. Combining these inequalities with (3.3) and (3.4), we obtain the statement of the lemma.  Note that if V is the three-element meet-semilattice that is not a lattice, then CoalL(V) is a lattice but there are singleton coalitions A0,A1,A2 ∈ CoalL(V) such that A1 ∧ A2 = A0 6⊆ A1 ∪ A2. Hence, the assumption that P is a chain is essential in Lemma 3.5. The following statement is taken from Cz´edli[10, Lemma 1]. Lemma 3.6 (Cz´edli[10, Lemma 1]). If X ≤ Y in a coalition lattice, then there exists an extensive map X → Y that acts identically on X ∩ Y . The height h(x) of an element x of a chain P is defined in the usual way: h(x) = k if and only if ↓x := {y ∈ P : y ≤ x} consists of k + 1 elements. The strength str(X) of a coalition X is defined to be X X str(X) := |X| + h(u) = |↓u|. u∈X u∈X DOUBLING TOLERANCES AND COALITION LATTICES 11

Note that for X ≤ Y in CoalL(P), we have that str(X) ≤ str(Y ). With these concepts, we can describe the covering relation in CoalL(P) as follows. Note that, as opposed to some parts of mathematics (far from lattice theory), here A ⊂ B means the conjunction of A 6= B and A ⊆ B. Lemma 3.7. Let P = (P ; <) be a finite chain with |P | ≥ 2, its smallest element is denoted by 0. Let A, B ∈ CoalL(P) such that A < B. Then A ≺ B in CoalL(P) if and only if the following two conditions hold. (i) str(B) = str(A) + 1; (ii) either A ⊂ B, or |A| = |B| = 1 + |A ∩ B|. Furthermore, for later reference, we note that (iii) if A ≺ B and A ⊂ B, then B = A ∪ {0}; (iv) if A ≺ B and |A| < |B|, then B = A ∪ {0}. Proof. Observe that (i) and (ii) together imply that A < B in CoalL(P). (3.5) Indeed, if A ⊂ B, then A < B is obvious. Assume that |A| = |B| = 1 + |A ∩ B|, then A = X ∪ {a} and B = X ∪ {b} with {a, b} ∩ X = ∅ and a 6= b. It follows from (i) that h(a) < h(b) and so a < b. Thus, {(a, b)} ∪ idX : A → B is an extensive map, and so A ≤ B. But A 6= B by (i), and we conclude that A < B, as required. This proves (3.5). Our next observation is that if A, B ∈ CoalL(P), then A < B implies that str(A) < str(B). (3.6) In order to see this, assume that A < B. Pick an extensive map ϕ: A → B. Since A 6= B, either ϕ is not surjective, or x < ϕ(x) for some x ∈ A, in addition to (∀y ∈ A)(y ≤ ϕ(y)), whereby str(A) < str(B) follows easily, proving (3.6). Combining (3.5) and (3.6), we obtain immediately that the conjunction of (i) and (ii) implies that A ≺ B in CoalL(P). Next, assume that A ≺ B. We are going to prove that (i) and (ii) hold. There are two subcases, depending on A ⊆ B or A 6⊆ B. First we deal with the case A ⊆ B. Since A 6= B, we have that A ⊂ B. In particular, we have already obtained that (ii) holds. If B\A had two distinct elements, x and y, then A ⊂ A∪{x} ⊂ A∪{x, y} ⊆ B would give that A < A ∪ {x} < A ∪ {x, y} ≤ B, contradicting A ≺ B. Hence, B \ A is a singleton {u}, that is, B = A∪{u}. We claim that u = 0. Suppose the contrary. Then A∪{0} < A∪{u} = B, either because 0 ∈ A and A∪{0} = A ⊂ A∪{u} = B, or because 0 ∈/ A and the extensive bijection idA ∪{(0, u)}: A∪{0} → A∪{u} = B is not the identity map since 0 6= u. Hence, A ≤ A ∪ {0} < A ∪ {u} = B, where the first inequality must be an equality since A ≺ B. Thus, 0 ∈ A and so A 6= B \{0}. Also, 0 ∈ B, because A ⊂ B, whence B \{0}= 6 B, whereby B \{0} ⊂ B and B \{0} < B. Since the map A → B \{0}, defined by 0 → u and x 7→ x for x ∈ A \{0}, is extensive, we obtain that A ≤ B \{0}. In fact, A < B \{0} since A 6= B \{0}. This inequality together with B \{0} < B contradict A ≺ B. Therefore, u = 0, B = A ∪ {0} ⊃ A, and str(B) = str(A) + h(0) + 1 = str(A) + 1. Thus, 3.7(i) and 3.7(ii) hold, as required. Second, still assuming that A ≺ B, we deal with the case A 6⊆ B, that is A \ B 6= ∅. We write A \ B in the form A \ B = {a1, a2, . . . , ak} where t ≥ 1 and the elements a1,..., at are pairwise distinct. Choose an extension map ϕ: A → B according to Lemma 3.6. Since ϕ acts identically on X := A ∩ B and ϕ is injective, 12 G. CZEDLI´ we have that a1 ≤ ϕ(a1) =: b1,..., at ≤ ϕ(at) =: bt are outside A ∩ B, so they are in B \ A. Since ai ∈/ B but bi ∈ B, we obtain that

ai < bi, for i = 1, . . . , t. (3.7)

With reference to the injectivity of ϕ again, we obtain that the elements b1,..., bt are pairwise distinct. Using the extensive maps

ϕeX ∪ {(a1, b1), (a2, a2), (a3, a3),..., (at, at)},

ϕeX ∪ {(b1, b1), (a2, b2), (a3, a3),..., (at, at)}, ...

ϕeX ∪ {(b1, b1), (b2, b2),..., (bt−1, bt−1), (at, bt)}, and the fact that X ∪ {b1, b2, . . . , bt} ⊆ B implies that X ∪ {b1, b2, . . . , bt} ≤ B, we obtain that  A = X ∪ {a1, . . . , at} < X ∪ {b1, a2, . . . , at}  < X ∪ {b1, b2, a3, . . . , at} < X ∪ {b1, b2, b3, a3, . . . , at} (3.8)  < ··· < X ∪ {b1, b2, . . . , bt} ≤ B.  Now we are in the position to conclude from A ≺ B and (3.8) that t = 1. In order to ease the notation, we let a := a1 and b := b1. Tailoring (3.8) to this new notation and t = 1, we have that A = X ∪ {a} < X ∪ {b} ≤ B. Hence, taking A ≺ B into account, B = X ∪ {b}. Using (3.7), we can summarize the situation as follows. A = X ∪ {a},B = X ∪ {b},X = A ∩ B, and a < b. (3.9) Clearly, 3.7(ii) is an immediate consequence of (3.9). We are going to show that h(b) = h(a) + 1, because then 3.7(i) will automatically follow from (3.9). For the sake of contradiction, suppose that h(b) 6= h(a) + 1. Then a < b yields that h(b) ≥ h(a) + 2. If the whole interval [a, b] is disjoint from X, then we can pick an element y such that a < y < b (that is, h(a) < h(y) < h(b) since P is a chain); this y is not in X and, witnessed by straightforward extensive functions extending idX , we have that A = X ∪ {a} < X ∪ {y} < X ∪ {b} = B, contradicting A ≺ B. Hence, X ∩ [a, b] 6= ∅, and so there is a unique smallest element z ∈ X such that a < z < b. Let C := (A \{z}) ∪ {b} = (X \{z}) ∪ {a, b} = (B \{z}) ∪ {a}. Clearly, z ∈ X = A ∩ B and z∈ / C give that A 6= C 6= B. The extension functions ( b, if x = z, A → C, defined by x 7→ and x, otherwise ( z, if x = a, C → B, defined by x 7→ x, otherwise, see Figure 4 for an illustration, indicate that A < C < B, contradicting A ≺ B. This shows that h(b) = h(a) + 1. Thus, with the exception of (iii) and (iv), the lemma is proved. Next, to prove (iii), assume that A ≺ B and A ⊂ B. We have already proved that (i) holds. We can write B in the form A ∪ {c1, . . . , ck}, where k ≥ 1 and A ∩ {c1, . . . , ck} = ∅. Using (i), we have that str(A) + 1 = str(B) = str(A) + k + h(c1) + ··· + h(ck), which implies that k = 1 and h(c1) = 0, that is, c1 = 0. Thus, B = A ∪ {0}, as required. Therefore, (iii) holds. DOUBLING TOLERANCES AND COALITION LATTICES 13

Figure 4. Illustration for h(b) ≥ h(a) + 2; A, B, and C consist of the black-filled elements

Finally, to prove (iv), assume that A ≺ B and |A| < |B|. Pick an extensive map ϕ: A → B. Since |A| < |B| and ϕ is injective, ϕ(A) ⊂ B, whereby ϕ(A) < B. Hence, A ≤ ϕ(A) < B and A ≺ B imply that A = ϕ(A). Thus, A = ϕ(A) ⊂ B, and (iii) applies. This proves (iv), and the proof of Lemma 3.7 is complete.  Lemma 3.8. Let P = (P ; <) be a finite chain with at least two elements, and let A, B ∈ CoalL(P). Let the smallest element and the unique atom of P be denoted by 0 and w, respectively. Let P0 be the subchain P 0 = P \{0} = ↑w with the inherited ordering. Then B covers A (in notation, A ≺ B) in the coalition lattice CoalL(P) if and only if one of the following three possibilities hold. (cov∗-1) 0 ∈/ A and B = A ∪ {0}. (cov∗-2) (i) Either 0 ∈/ A ∪ B, B 6= A ∪ {w}, and A ≺ B in CoalL(P0), (ii) or 0 ∈ A ∩ B, B 6= A ∪ {w}, and A \{0} ≺ B \{0} in CoalL(P0). (cov∗-3) 0 ∈ A, w∈ / A, and B = (A \{0}) ∪ {w}. Proof. It is trivial to see that each of (cov∗-1), (cov∗-2), and (cov∗-3) (3.10) implies that A < B in CoalL(P). Next, we claim that for every X ∈ CoalL(P), the only extensive X → X map  (3.11) is the identity map idX : X → X, defined by x 7→ x. We show this by induction on |X|. For |X| ≤ 1, (3.11) is clear. For |X| > 1 and an arbitrary extensive map ϕ: X → X, the ϕ-image of the largest element b of X is necessarily b, because b ≤ ϕ(b). Let Y =: X \{b}. The restriction ϕeY of ϕ to Y is a Y → Y map by injectivity, and so ϕeY : Y → Y is an extensive map. Since ϕeY = idY by the induction hypothesis, we obtain that ϕ = idX , proving (3.11). By Lemma 3.7 and (3.10), (cov∗-1) implies that A ≺ B. Assume (cov∗-2i). Then, since A ≺ B in CoalL(P0), Lemma 3.7 yields that 3.7(i) and 3.7(ii) hold for A and B over P0. Since B 6= A ∪ {w} and w is the smallest element of P0, 3.7(iii) excludes that A ⊂ B. Hence, 3.7(ii) leads to |A| = |B| = 1 + |A ∩ B|; this holds not only over P0 but also over P. That is, 3.7(ii) holds for A and B over P. For 0 0 x ∈ P , hP (x) = 1 + hP0 (x). Hence, for every X ∈ CoalL(P ), X X  strP (X) = |X| + hP (u) = |X| + (hP0 (u) + 1)  u∈X u∈X  X (3.12) = |X| + |X| + h 0 (u) = |X| + str 0 (X)  P P  u∈X 14 G. CZEDLI´

Thus, using (3.12), |A| = |B|, and that 3.7(i) holds for A and B in over P0,

strP (B) = |B| + strP0 (B) = |A| + strP0 (A) + 1 = strP (A) + 1, that is, 3.7(i) holds for A and B in over P. Hence, (3.10) and Lemma 3.7 imply that A ≺ B in CoalL(P), as required. Next, when assuming (cov∗-2ii), we are going to reduce the task to (cov∗-2i), which has just been settled. Namely, observe that (cov∗-2i) holds for A0 := A \{0} and B0 := B \{0}. Apart from slight notational changes, we derived from this situation that 3.7(ii) holds for A0 and B0 with |A0| = |B0| = 1 + |A0 ∩ B0| and that 3.7(i) also holds for A0 and B0 over P (and so A0 ≺ B0 in CoalL(P) but this is not relevant at this moment). These two facts imply that 3.7(ii) and 3.7(i) holds for A = A0 ∪ {0} and B = B0 ∪ {0} over P, whereby A ≺ B in CoalL(P) by (3.10) and Lemma 3.7, as required. Finally, if (cov∗-3), then |A| = |B| = |A ∩ B| + 1 and str(B) = str(A) + h(w) − h(0) = str(A) + 1, and so (3.10) together with Lemma 3.7 yield the required A ≺ B in CoalL(P). We have seen that the disjunction of (cov∗-1), (cov∗-2), and (cov∗-3) is a sufficient condition of A ≺ B. In order the see that the above-mentioned disjunction is a necessary condition, the rest of the proof assumes that A ≺ B in CoalL(P). Note in advance that then our assumption, A ≺ B, excludes that B = A ∪ {w}, (3.13) since otherwise B 6= A would give that w∈ / A and so str(B) = str(A) + 1 + h(w) = str(A) + 2, which would contradict Lemma 3.7(i). By Lemma 3.7, 3.7(i) and 3.7(ii) hold for A and B over P. According to the containment of 0 in A and B, there are four cases to consider. First, assume that 0 ∈/ A and 0 ∈/ B. Then Lemma 3.7(iv) excludes that |A| < |B|, whence 3.7(ii) imply that |A| = |B| = 1 + |A ∩ B|, which holds also over P0. In particular, 3.7(ii) holds over P0. Using 3.7(i) over P and |A| = |B|, and computing by (3.12), we obtain the validity of 3.7(i) over P0. Hence, Lemma 3.7 gives that A ≺ B in CoalL(P0). Thus, taking (3.13) also into account, we obtain that (cov∗-2i) holds. Second, assume that 0 ∈ A and 0 ∈ B, and let A0 := A \{0} and B0 := B \{0}. Lemma 3.6 gives easily that A0 < B0 in CoalL(P). From A 6= B and Lemma 3.7(iii), we conclude that A 6⊂ B. Thus, Lemma 3.7(ii) gives that |A| = |B| = 1 + |A ∩ B|. This implies that |A0| = |B0| = 1 + |A0 ∩ B0|, that is, 3.7(ii) holds for A0 and B0. Since str(B) = str(A) + 1 by 3.7(i), we obtain that 3.7(i) holds also for A0 and B0. Combining these facts with Lemma 3.7, we obtain that A0 ≺ B0 in CoalL(P). The previous paragraph has shown that this yields the validity of (cov∗-2i) for A0 and B0. This fact implies trivially that (cov∗-2ii) holds for A and B. Third, assume that 0 ∈ A but 0 ∈/ B. Then A 6⊂ B, whereby Lemma 3.7(ii) leads to |A| = |B| = |A ∩ B| + 1. That is, A = X ∪ {a} and B = X ∪ {b} with a 6= b, {a, b} ∩ X = ∅, a∈ / B, and b∈ / A. Since both 0 and a are in the singleton set A \ X, we have that a = 0, and so Lemma 3.7(i) yields that str(X) + 1 + h(b) = str(B) = str(A) + 1 = str(X) + 1 + h(0) + 1 = str(X) + 2. Hence, h(b) = 1, that is, b = w. Consequently, B = (A \{0}) ∪ {w}, w = b∈ / A, and (cov∗-3) holds. Fourth, assume that 0 ∈/ A but 0 ∈ B. If we had that |A| = |B| = |A∩B|+1, then we would have that A = X ∪ {a} and B = X ∪ {b} with a 6= b and {a, b} ∩ X = ∅, DOUBLING TOLERANCES AND COALITION LATTICES 15 whereby b = 0 and a 6= 0 would give that str(A) = str(X) + 1 + h(a) > str(X) + 1 + h(b) = str(B), contradicting Lemma 3.7(i). Hence, A ⊂ B by Lemma 3.7(ii), B = A ∪ {0} by Lemma 3.7(iii), and (cov∗-1) holds. We have seen that whenever A ≺ B in CoalL(P), then the disjunction of (cov∗-1), ∗ ∗ (cov -2), and (cov -3) holds. This completes the proof of Lemma 3.8.  As a preparation, let us recall the following useful result of Gr¨atzer[15]. Lemma 3.9 (Gr¨atzer[15]). Let Θ be an equivalence relation on a finite lattice L such that the Θ-blocks are intervals. Then Θ is a congruence if and only if ∀x, y, z ∈ L, if z ≺ x, z ≺ y, and (z, x) ∈ Θ, then (y, x ∨ y) ∈ Θ, and dually. Proof of Lemma 3.4. By a doubling congruence we mean a transitive doubling tol- erance, that is, a doubling tolerance that happens to be a congruence. Note that a doubling tolerance is a doubling congruence (3.14) if and only if its blocks are pairwise disjoint. We prove the lemma by induction on the size |P | of the chain P. The base of the induction, |P | = 2, is trivial, whereby the rest of the proof is devoted to the induction step. With the notation L := CoalL(P0) and M := CoalL(P), we know from the induction hypothesis that L is distributive. We need to show that T is a ∼ doubling congruence on L and [L ∗2 T ] = M; then Theorem 2.1 will immediately imply that M is also distributive. From (3.14) and the definition of T in (3.1), it follows that T is an equivalence relation and each of o (3.15) its blocks is a two-element interval. Furthermore, the distributivity (in fact, the modularity) of L implies that whenever z ≺ x and z ≺ y in L, then (z, x, y, x ∨ y) is a covering square in L, and dually. Therefore, by Lemma 3.9, in order to conclude that T is a doubling congruence, it suffices to show that if T collapses an edge of a covering square, then (3.16) it collapses the opposite edge of the square. Let (A = B ∧ C,B,C,D = B ∨ C) be a covering square in L = CoalL(P0). First, assume that a lower edge, say, A ≺ B, is collapsed by T . This means that B = A ∪ {w} and w∈ / A. Since A ⊂ B and Lemma 3.7(iii) allows only one X such that A ≺ X and A ⊂ X, it follows from Lemma 3.7(ii) that, in addition to B = A∪{w}, we have that |A| = |C|; note that w, the smallest element of P0, plays the role of 0 in Lemma 3.7. Since w is already in B, (ii) and (iii) of Lemma 3.7 give that |B| = |D|. But then |C| = |A| < |B| = |D|, C ≺ D, and Lemma 3.7(iv) yield that D = C ∪ {w}, whereby (C,D) ∈ T . So, T “spreads” from a lower edge to the opposite upper edge. Second, assume that an upper edge, say, C ≺ D, is collapsed by T , that is, D = C ∪ {w} and w∈ / C; the argument is almost the same as above. Namely, if we had |B| < |D|, then Lemma 3.7(iv) would give that D = B ∪ {w} and we would obtain that B = D \{w} = C, a contradiction. Hence, |B| = |D|. Since |A| < |C| would contradict w∈ / C by Lemma 3.7(iv), |A| = |C|. Hence, |A| = |C| < |D| = |B|, which together with A ≺ B and Lemma 3.7(iv) yield that B = A ∪ {w}, whereby A ≺ B is collapsed by T , as required. We have seen the validity of (3.16), whereby we have shown that T is a doubling congruence on L. 16 G. CZEDLI´

Next, we define the following map ( A, if k = 0, γ :[L ∗2 T ] → M by (A, k) 7→ A ∪ {0}, if k = 1, and we are going to show that γ is a lattice isomorphism. Since γ is trivially a bijection, it suffices to show that γ is an order-isomorphism. Furthermore, since orderings on finite posets are determined by the corresponding covering relations, our task reduces to proving that x ≺ y in [L ∗2 T ] if and only if γ(x) ≺ γ(y) in M. The covering pairs in [L ∗2 T ] and those in M are described by (cov-1)–(cov-3) from the proof of Theorem 2.1 and by (cov∗-1)–(cov∗-3) from Lemma 3.8, respectively. Therefore, it suffices to prove that for any (A, i) and (B, j) in [L ∗2 T ] and for any ` ∈ {1, 2, 3}, (cov-`) holds for (A, i) and (B, j) if and only  (3.17) if (cov∗-`) holds for γ(A, i) and γ(B, j).

Assume that (cov-1) holds for (A, i) and (B, j) in [L ∗2 T ]. That is, (A, i) = (A, 0) and (B, j) = (A, 1). Hence. γ(A, i) = A and γ(B, j) = A ∪ {0}, whereby γ(A, i) and γ(B, j) satisfy (cov∗-1). Conversely, assume that γ(A, i) and γ(B, j) satisfy (cov∗-1). Then 0 ∈/ γ(A, i) and 0 ∈ γ(B, j) = γ(A, i) ∪ {0} imply that i = 0 and j = 1. Hence, by the definition of γ, we have that A ∪ {0} = γ(A, 0) ∪ {0} = γ(B, 1) = B ∪ {0}, whereby A = B. Thus, (A, i) = (A, 0) and (B, j) = (A, 1) satisfy (cov-1), as required; this settles (3.17) for ` = 1. Next, assume that (cov-2) holds for (A, i) and (B, j) in [L ∗2 T ]. That is, A ≺ B in L and i = j, but A 6≺T B. Since A 6≺T B, we have that B 6= A ∪ {w}. For the case i = j = 1, note that B 6= A ∪ {w} implies that B ∪ {0} 6= (A ∪ {0}) ∪ {w}. Thus, no matter if i = j is 0 or 1, it follows that (cov∗-2) holds for γ(A, i) and γ(B, j). Conversely, assume that (cov∗-2) holds for γ(A, i) and γ(B, j); there are two cases depending on the containment of 0 in γ(A, i). First, assume that 0 ∈/ γ(A, i). Then (cov∗-2ii) is excluded, so (cov∗-2i) holds for γ(A, i) and γ(B, j). Hence 0 ∈/ γ(B, j), i = j = 0, A = γ(A, 0) ≺ γ(B, 0) = B in L, and B 6= A ∪ {w}. Let us summarize for later reference that i = j, A ≺ B in L, and B 6= A ∪ {w}. (3.18) We are going to show that (3.18) implies that (cov-2) holds for (A, i) and (B, j). Since A ≺ B excludes that A = B, the equality A ∪ {w} = B ∪ {w} would only be possible if we had that B = A ∪ {w}, excluded above, or A = B ∪ {w} ⊃ B, excluded by A < B. Thus A ∪ {w}= 6 B ∪ {w}, that is, (A, B) ∈/ T and A 6≺T B. Therefore, (cov-2) holds for (A, i) and (B, j), as required. Second, assume that 0 ∈ γ(A, i). Now (cov∗-2i) is excluded, so (cov∗-2ii) holds for γ(A, i) and γ(B, j). In particular, 0 ∈ γ(B, j). Hence, i = j = 1, γ(A, i) = A ∪ {0}, and γ(B, j) = B ∪ {0}. Thus, the validity of (cov∗-2ii) for these two sets gives that A = (A∪{0})\{0} ≺ (B∪{0})\{0} = B in L and B∪{0}= 6 B∪{0}∪{w}. Since this non-equality gives that B 6= A∪{w}, (3.18) is fulfilled. We already know that (3.18) implies that (cov-2) holds for (A, i) and (B, j). Therefore, we have shown (3.17) for ` = 2. Finally, assume that (cov-3) holds for (A, i) and (B, j) in [L ∗2 T ]. That is, i = 1, j = 0, and, furthermore, A ≺T B, which gives that w∈ / A and B = A ∪ {w}. Hence, γ(A, i) = A ∪ {0} and γ(B, j) = B, and they clearly satisfy (cov∗-3) since DOUBLING TOLERANCES AND COALITION LATTICES 17

(γ(A, i)\{0})∪{w} = ((A∪{0})\{0})∪{w} = A∪{w} = B = γ(B, j). Conversely, assume that (cov∗-3) holds for γ(A, i) and γ(B, j). Then 0 ∈ γ(A, i) and 0 ∈/ γ(B, j) yield that i = 1 and j = 0. Furthermore, w∈ / γ(A, i) = γ(A, 1) = A ∪ {0} give that w∈ / A, and we also have that B = γ(B, j) = (γ(A, 1) \{0}) ∪ {w} = ((A ∪ {0}) \{0}) ∪ {w} = A ∪ {w}. Hence, B ∪{w} = A∪{w} and (A, B) ∈ T . Combining this with A ⊂ B and (3.15), we obtain that A ≺T B. Thus, (cov-3) holds for (A, i) = (A, 1) and (B, j) = (B, 0) in [L ∗2 T ]. This completes the proof of (3.17) and that of Lemma 3.4.  References [1] J.M. Alonso-Meijide, M. Alvarez-Mozos,´ M.G. Fiestras-Janeiro, and A. Jim´enez-Losada: Some structural properties of a lattice of embedded coalitions. Int. J. General Systems 46, 123–143 (2017) DOI:10.1080/03081079.2017.1297431 [2] H.-J. Bandelt: Tolerance relations on lattices. Bull. Austral. Math. Soc. 23, 367–381 (1981) [3] A. Bj¨orner,P. H. Edelman, G. M. Ziegler: Hyperplane arrangements with a lattice of regions. Discrete Comput. Geom. 5, 263–288 (1990) [4] I. Chajda: Algebraic Theory of Tolerance Relations. Palack´yUniversity Olomouc, 1991; https://www.researchgate.net/publication/36797871_Algebraic_Theory_of_Tolerance_Relations [5] I. Chajda, G. Cz´edli, and R. Halaˇs:Independent joins of tolerance factorable varieties. Alge- bra Universalis 69, 83–92 (2013) [6] I. Chajda, G. Cz´edli,and I. G. Rosenberg: On lattices whose ideals are all tolerance kernels. Acta Sci. Math. (Szeged) 61, 23–32 (1995) [7] I. Chajda and B. Zelinka: Tolerance relations on lattices. Casop.ˇ Pˇestov. Mat. 99, 394–399 (1974). [8] G. Cz´edli:Factor lattice by tolerances. Acta Sci. Math. (Szeged) 44, 35–42 (1982) [9] G. Cz´edli:A Horn sentence in coalition lattices. Acta Math. Hungarica 72, 99–104 (1996) [10] G. Cz´edli: Jordan-H¨oldercondition with subsemilattices of coalition lattices, Contributions to General Algebra 16 (Proc. Conf Dresden 2004, AAA68, and Summer School 2004), Verlag Johannes Heyn, Klagenfurt 2005, 55–62. MR2166945 [11] G. Cz´edliand G. Gr¨atzer: Lattice tolerances and congruences. Algebra Universalis 66, 5–6 (2011) [12] G. Cz´edliand Gy. Poll´ak:When do coalitions form a lattice?. Acta Sci. Math. (Szeged), 60, 197–206 (1995) [13] G. Cz´edli,B. Larose, and Gy. Poll´ak:Notes on coalition lattices. Order 16, 19–29 (1999) [14] Gr¨atzer,G.: Lattice Theory: Foundation. Birkh¨auser,Basel (2011) [15] G. Gr¨atzer: A technical lemma for congruences of finite lattices. Algebra Universalis 72, 53–55 (2014) [16] J. Grygiel and S. Radeleczki: On the tolerance lattices of tolerance relations. Acta Math. Hungar. 141, 220–237 (2013) [17] J. Jakub´ık: Modular lattices of locally finite length. Acta Sci. Math. (Szeged), 37, 79–82 (1975) [18] Katsuya Nakano, Shun Shiramatsu, Tadachika Ozono, and Toramatsu Shintani: Coalition Lattice: A Data Structure considering Robustness for Robust Coalition Structure Gener- ation Problem. Proceedings of the 3rd International Conference on Industrial Application Engineering (2015) DOI: 10.12792/iciae2015.009 [19] M. Kindermann: Uber¨ die Aquivalenz¨ von Ordnungspolynomvollst¨andigkeit und Toleranze- infachheit endlicher Verb¨ande.Contributions to general algebra (Proc. Klagenfurt Conf., Klagenfurt, 1978), 145–149, Heyn, Klagenfurt, 1979 [20] Tuomas Sandholm, Kate Larson, Martin Andersson, Onn Shehory, and Fernando Tohm: Coalition structure generation with worst case guarantees. Artificial Intelligence 111 (1999) 209–238

University of Szeged, Bolyai Institute, Szeged, Aradi vertan´ uk´ tere 1, Hungary 6720 E-mail address: [email protected] URL: http://www.math.u-szeged.hu/~czedli/